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The x coordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?



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sauravshakya
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9y^2=x^3
d(9y^2)/dx =d(x^3)/dx
18y dy/dx =3x^2
dy/dx =x^2/(6y)

sauravshakya
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The gradient of tangent is x^2/(6y)
So, gradient of normal is 6y/(x^2)

sauravshakya
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NOW, let xintercept of the normal be a then yintercept is also a.

sauravshakya
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Gradient of normal = (a0)/(0a)
6y/(x^2) =1

sauravshakya
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Now,dw:1359201825424:dw
and solve for x

sauravshakya
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dw:1359201971611:dw

sauravshakya
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U need to solve that

Yahoo!
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dw:1359202095063:dw

sauravshakya
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9y^2=x^3

Yahoo!
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6y/(x^2) =1......then wat did u do ?

sauravshakya
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Substitute y interms of x